Math 25: Advanced Calculus
UC Davis, Spring 2011
Math 25 — Solutions to Homework Assignment #1
Remember throughout that there may be more than one correct way to prove
any given result.
Solutions
1. Let
n
be a natural number, divisible both by 5 and by 3. Because
n
is
divisible by 5, there exists some natural number
m
such that
n
= 5
m
.
Since
n
is also divisible by 3,
n
3
=
5
m
3
must be natural. Now, we use the
twin facts that any natural multiple of a natural number is natural, and
that any (nonnegative) diﬀerence between natural numbers must also
be natural. Using the ﬁrst, 2
m
=
6
m
3
is natural; so, using the second,
6
m
3

5
m
3
=
m
3
is natural. By substitution, then,
(
n
5
)
3
=
n
15
is a natural
number, and hence
n
is divisible by 15.
2. We disprove the claim directly, by oﬀering a counterexample. Let
n
=
24. Then
n
6
= 4 and
n
8
= 3, so that
n
is divisible both by 6 and by 8.
However,
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 Winter '11
 Wiley
 Differential Equations, Calculus, Equations, Prime number, 2m, 5m

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